What is the integral of 16tan^2(x)sec(x) ?

The integral of 16tan^2(x)sec(x) is 16(1/2 sec(x)tan(x)-1/2 ln|tan(x)+sec(x)|)+C

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integral of 16tan^2(x)sec(x)

\int 16 tan^{2} (x) sec (x) d x

16 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

Solution steps

\int 16 tan^{2} (x) sec (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int tan^{2} (x) sec (x) d x

Rewrite using trig identities

= 16 \cdot \int (- 1 + sec^{2} (x)) sec (x) d x

Expand

(- 1 + sec^{2} (x)) sec (x) : - sec (x) + sec^{3} (x)

= 16 \cdot \int - sec (x) + sec^{3} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (- \int sec (x) d x + \int sec^{3} (x) d x)

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 16 (- ln ∣ tan (x) + sec (x) ∣ + \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Simplify

- ln ∣ tan (x) + sec (x) ∣ + \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣ : \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 16 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 16 (\frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

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What is the integral of 16tan^2(x)sec(x) ?
The integral of 16tan^2(x)sec(x) is 16(1/2 sec(x)tan(x)-1/2 ln|tan(x)+sec(x)|)+C